Example 5 1 worked on the board
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Example 5.1 Worked on the Board! PowerPoint PPT Presentation


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Example 5.1 Worked on the Board!. Find the gravitational potential Φ inside & outside a spherical shell, inner radius b , outer radius a . (Like a similar electrostatic potential problem!)

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Example 5.1 Worked on the Board!

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Example 5 1 worked on the board

Example 5.1Worked on the Board!

  • Find the gravitational potential Φ inside & outside a spherical shell, inner radius b, outer radius a. (Like a similar electrostatic potential problem!)

  • This is an important & fundamental problem in gravitational theory! If you understand this, you understand the gravitational potential concept!(& probably the electrostatic potential concept as well!) .

  • Could approach the spherical shell using forces directly (Prob. 5.6), but its MUCH easier with the potential!


Example 5 1 worked on the board

  • Find Φ inside & outside a spherical shell of mass M, mass density ρ,inner radius b & outer radius a.

    Φ = -G∫[ρ(r)dv/r]. Integrate over V.The difficulty, of course, is

    properly setting up the integral!If properly set up, doing it is easy!


Example 5 1 worked on the board

Recall Spherical Coordinates


Example 5 1 worked on the board

Outline of Calculation!


Example 5 1 worked on the board

Summary of ResultsM(4π)ρ(a3 - b3)

outside the shell

R > a, Φ = -(GM)/R (1)

The same as ifM were a point mass at the origin!

completely inside the shell

R < b, Φ = -2πρG(a2 - b2) (2)

Φ = constant, independent of position.

within the shell

b  R  a, Φ = -4πρG[a2- (b3/R) - R2] (3)

  • Also, Φis continuous!

     If R  a, (1) & (3) are the same! If R  b, (2) & (3) are the same!


Example 5 1 worked on the board

  • These results arevery important,especially

    those forR > a, Φ = -(GM)/R

     This says: The potential at any point outside

    a spherically symmetric distribution of matter(shell or solid; a solid is composed of infinitesimally thick shells!)is independent of the size of the distribution & is the same as that for a point mass at the origin.

    To calculate the potential for such a distribution we can consider all mass to be concentrated at the center.


Example 5 1 worked on the board

  • Also, the results are very important for

    R < b, Φ = -2πρG(a2 - b2)

    The potential is constant anywhere

    inside a spherical shell.

    The force on a test mass m

    inside the shell is 0!


Example 5 1 worked on the board

  • Given the results for the potential Φ,we can compute the GRAVITATIONAL FIELD g inside, outside & within the spherical shell: g  - Φ

  • Φdepends on R only  g is radially directed

    g = g er = - (dΦ/dR)er [M(4π)ρ(a3 - b3)]

    outside the shellR > a, g = - (GM)/R2The same as ifM were a point mass at the origin!

    completely inside the shell R < b, g = 0

    Since Φ = constant, independent of position.

    within the shell

    b  R  a, g = (4π)ρG[(b3/R2) - R]


Example 5 1 worked on the board

  • Plots of the

    potential

    Φ & the field g

    inside, outside &

    within a

    spherical shell.

    g  - Φ

    g = - (dΦ/dR)

 Φ = constant

Φ = -(GM)/R

g = - (GM)/R2

g = 0


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